Question

A body is projected with a speed of 40m/s at angle of projection of 30°. Find :
1.) The maximum height it reaches
2.) the time of ascent
3.) the time of descent
4.) the range of motion
5) its velocity at t=1s,t=3s.
6.) its position at t=1s andt=3s.
7.) the velocity with which it strikes the ground.​

Answer 1

Explanation:

u = 40m/s

theta = 30°

1) Max height

H = u^2 sin^2 theta /2g

= 1600 × 1/4/20

= 400/20

= 20 m

2) Time(ascent)

T = 2u sin theta /g

= 80 × 1/2 /10

= 4sec

3) Time (decent)

time of ascent = time of decent= 4sec

4) Range

R = u^2 sin2theta /g

= 800√3/10

= 80√3 m

5)its velocity at t=1s,t=3s.

v = u +at

= 40 + 10 × 2

= 60

v = u+at

= 40 + 10×3

= 70

6)position at t=1s andt=3s.